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Full text of "Treatise On Analysis Vol-Ii"

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canonical factorization E -> E/F2 ~* F1? where y\ is a bijection. To say that E
is the topological direct sum of F1 and F2 signifies that j\ is continuous
(12.10.6), or equivalently that the inverse bijection, which is the restriction to
F! of the canonical mapping n2 : E -> E/F2 (and which is always continuous)
is bicontinuous.

(12.13.2) (i) Let E be a topological vector space and let H be a hyperplane in
E, with equation f(x) = 0. Then H is closed in E if and only if the linear form f is
continuous on E.

(ii)   Let E be a Hausdorff topological vector space over R (resp. C) of
finite dimension n. //'(aI.)1^l-^n is a basis ofE, the mapping

o/R" (resp. Cn) onto E is an isomorphism of topological vector spaces.

(iii) Let E be a topological vector space, M a closed vector subspace ofE,
and F a finite-dimensional vector subspace ofE. Then M 4- F is closed in E. In
particular, if E is Hausdorff, every finite-dimensional vector subspace of E is
closed in E.

(iv) Let E be a topological vector space and let M be a closed vector sub-
space 0/finite codimension in E. Then every algebraic supplement N ofM in E
is a topological supplement.

We shall prove (ii) first. Since u is continuous and bijective, we have to
show that u is bicontinuous, or equivalently (considering the inverse images
under u of open sets in E) that if O is a Hausdorff topology compatible with
the vector space structure of Rn (resp. C"), and coarser than the product
topology D0, then O = D0. Put \\x\\ = sup|^| for x = (f) e R" (resp. C") and
consider a ball B : ||jc|| < r with respect to this norm. The sphere S : ||x|| = r
of the same radius is compact in R" (resp. C") for the product topology 50
((3.17.6), (3.20.16), and (3.17.3)). For each z e S, there exists a neighborhood
V(z) of z and a neighborhood W(z) of 0, in the topology O, which do not
intersect. Since V(z) is also a neighborhood of z in the topology O0, there
exists finitely many points z, e S such that the union of the V(zt-) contains S.
Now let U be a balanced neighborhood of 0 in the topology O (12.13.1), con-
tained in the intersection of the neighborhoods W(zj). IfxeU were such that
\\x\\ ^ r, then the point rx/ \\x\\ would belong to both S and U, which is
absurd. Hence U c B, which establishes our assertion (12.2.1).

Part (iv) of the proposition is an immediate consequence of (ii). For the
set {0} = M n N is closed in N, and therefore N is Hausdorff (12.8.3). Sinceor arbitrary topological vector spaces.